Let a tangent to the curve $$y^2 = 24x$$ meet the curve $$xy = 2$$ at the points $$A$$ and $$B$$. Then the midpoints of such line segments $$AB$$ lie on a parabola with the

We need to find the locus of midpoints of line segments $$AB$$ where a tangent to $$y^2 = 24x$$ meets $$xy = 2$$.

For the parabola $$y^2 = 24x$$, we have $$4a = 24$$ so that $$a = 6$$, and its tangent in slope form is given by $$y = mx + \frac{6}{m}$$.

The curve $$xy = 2$$ can be written as $$y = \frac{2}{x}$$. Substituting this into the tangent equation yields

$$\frac{2}{x} = mx + \frac{6}{m},$$

which simplifies to

$$2 = mx^2 + \frac{6x}{m}$$

and then to the quadratic

$$m^2x^2 + 6x - 2m = 0.$$

Let the points of intersection be $$A = (x_1,y_1)$$ and $$B = (x_2,y_2)$$. By Vieta’s formulas, the roots of the quadratic satisfy

$$x_1 + x_2 = \frac{-6}{m^2},\qquad x_1 x_2 = \frac{-2}{m}.$$

If $$(h,k)$$ is the midpoint of $$AB$$, then

$$h = \frac{x_1 + x_2}{2} = \frac{-3}{m^2}.$$

Since $$y_i = mx_i + \frac{6}{m}$$ for each intersection point, the midpoint’s $$y$$-coordinate is

$$k = \frac{y_1 + y_2}{2} = m\cdot h + \frac{6}{m} = m\cdot\frac{-3}{m^2} + \frac{6}{m} = \frac{-3}{m} + \frac{6}{m} = \frac{3}{m}.$$

From $$k = \frac{3}{m}$$ we have $$m = \frac{3}{k}$$. Substituting into the expression for $$h$$ gives

$$h = \frac{-3}{m^2} = \frac{-3}{\bigl(\tfrac{3}{k}\bigr)^2} = \frac{-3k^2}{9} = \frac{-k^2}{3}.$$

Hence the coordinates of the midpoint satisfy

$$k^2 = -3h,$$

or in the usual $$(x,y)$$-form,

$$y^2 = -3x.$$

This is a left-opening parabola with equation $$y^2 = -3x$$, for which $$4a = 3$$ so that $$a = \tfrac{3}{4}$$ and the vertex is at the origin. Its directrix is

$$x = \frac{3}{4},$$

or equivalently $$4x = 3$$, and the length of the latus rectum is $$4a = 3$$.

The answer is Option 1: directrix $$4x = 3$$.